Sex Linkage
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You should be comfortable with Mendelian monohybrid crosses, the concept of dominant and recessive alleles, and the basic structure of human and Drosophila sex chromosomes (XX female, XY male).
Why this matters
Sex linkage explains why certain genetic disorders and traits show a striking bias toward one sex in inheritance patterns.
While most autosomal genes are inherited equally by sons and daughters, genes located on the X chromosome behave differently because males possess only one copy. This single-copy status means that males express whatever allele they inherit on their X chromosome, without the masking effect of a second allele. Consequently, recessive X-linked traits, such as red-green colour blindness or haemophilia, appear far more frequently in males than in females. Understanding this mechanism is crucial for genetic counselling, as it allows us to predict the risk of transmission from carrier mothers to their sons and daughters with high accuracy.
The Asymmetry of the X Chromosome
To understand sex linkage, we must first recognise the fundamental difference in chromosome number between the sexes. In humans and many other mammals, females are homogametic, possessing two X chromosomes, while males are heterogametic, possessing one X and one Y chromosome. The Y chromosome is relatively small and carries few genes, most notably the SRY gene which triggers testis development. The X chromosome, however, is large and carries hundreds of genes, including those responsible for vision, blood clotting, and muscle function. Because males have only one X chromosome, they are hemizygous for all X-linked genes. This means there is no second allele to potentially mask a recessive one. If a male inherits a defective allele for an X-linked gene, he will express the trait or disorder. In contrast, a female must inherit two defective alleles—one from each parent—to express a recessive X-linked trait. If she inherits one defective allele and one normal allele, she is a carrier. She typically does not show the phenotype because the normal allele produces sufficient functional protein to maintain health. This biological asymmetry is the root of the characteristic inheritance pattern where affected fathers pass the trait to all their daughters (who become carriers) but none of their sons, while carrier mothers pass the trait to half their sons (who are affected) and half their daughters (who become carriers).
Genotypic and Phenotypic Ratios
When we analyse a cross between a heterozygous carrier female and a normal male, the outcomes are distinct for each sex. Let us denote the normal allele as \( X^{A} \) and the recessive disease allele as \( X^{a} \). The female parent is \( X^{A}X^{a} \) and the male parent is \( X^{A}Y \). During gamete formation, the female produces two types of eggs in equal proportions: \( X^{A} \) and \( X^{a} \). The male produces two types of sperm: \( X^{A} \) and \( Y \). Fertilisation is a random event, leading to four possible zygotic combinations. Two of these combinations result in females: \( X^{A}X^{A} \) (normal) and \( X^{A}X^{a} \) (carrier). The other two result in males: \( X^{A}Y \) (normal) and \( X^{a}Y \) (affected). Therefore, among the female offspring, 50% are phenotypically normal but carriers, and 50% are completely normal. Among the male offspring, 50% are phenotypically normal, and 50% are affected by the trait. This 1:1 ratio of affected to unaffected males is a hallmark of X-linked recessive inheritance. It is important to note that the overall population ratio of affected individuals is not 25%, because the expression is sex-dependent. If we look at the total offspring, 25% are affected males, 25% are carrier females, 25% are normal females, and 25% are normal males. This specific distribution is a direct consequence of the hemizygous state of the male.
Transmission Patterns in Pedigrees
The pattern of transmission in a family tree provides strong evidence for X-linkage. A key diagnostic feature is the absence of male-to-male transmission. An affected father has the genotype \( X^{a}Y \). He passes his Y chromosome to all his sons and his \( X^{a} \) chromosome to all his daughters. Therefore, none of his sons can inherit the disease allele from him, as they receive the Y chromosome. However, all his daughters will receive the \( X^{a} \) chromosome and thus become carriers (assuming the mother contributes a normal \( X^{A} \)). This contrasts sharply with autosomal recessive inheritance, where an affected father can pass the allele to his sons. Another pattern to observe is that the trait often skips generations. An affected male may have a mother who is a carrier, who in turn may have a father who was affected. This creates a 'zig-zag' pattern through the pedigree. In contrast, if a mother is affected (\( X^{a}X^{a} \)), she must pass an \( X^{a} \) to all her children. Consequently, all her sons will be affected, and all her daughters will be at least carriers. This high penetrance in the offspring of an affected mother is a strong indicator of X-linked recessive inheritance. These patterns allow geneticists to distinguish sex-linked traits from autosomal ones with high confidence, even in small family samples.
Biological Consequences and X-Inactivation
One might wonder why females, who have two X chromosomes, do not have twice the gene dosage of males. This is resolved by the process of X-inactivation, also known as Lyonisation. Early in embryonic development, one of the two X chromosomes in each female cell is randomly inactivated and condensed into a Barr body. This ensures that females and males have a similar level of expression for X-linked genes. However, because the inactivation is random, a female carrier is a mosaic of cells. Some cells express the normal allele, while others express the mutant allele. In most cases, the normal cells compensate for the mutant ones, and the female remains phenotypically normal. However, in rare cases, if the X chromosome carrying the normal allele is inactivated in a critical proportion of cells in a specific tissue, the female may exhibit a mild form of the disease. This phenomenon, known as skewed X-inactivation, explains why some carrier females for haemophilia or Duchenne muscular dystrophy may show mild symptoms. It also highlights the complexity of sex-linked inheritance, where the genotype does not always predict the phenotype with absolute certainty due to epigenetic regulation.
The concept
Sex linkage
Inheritance of genes carried on the sex chromosomes.
A common misconception is that because males are hemizygous, they are more likely to be affected by X-linked dominant disorders. In reality, X-linked dominant disorders affect males and females equally in terms of inheritance probability from a carrier parent, but males are often more severely affected because they lack a second, potentially normal, copy of the gene to buffer the effect of the mutation. For example, in X-linked dominant hypophosphataemia, males may have more severe rickets than carrier females. This severity difference is due to the lack of genetic redundancy in males, not a higher probability of inheritance. C
Worked example
Problem. A woman with normal vision (genotype \( X^{A}X^{a} \)) marries a man with red-green colour blindness (genotype \( X^{a}Y \)). Calculate the probability that their first child is a colour-blind daughter.
- Identify the parental genotypes: Mother is \( X^{A}X^{a} \), Father is \( X^{a}Y \).
- Determine the possible gametes: Mother produces \( X^{A} \) and \( X^{a} \) eggs; Father produces \( X^{a} \) and \( Y \) sperm.
- List the possible offspring genotypes: \( X^{A}X^{a} \) (normal vision female), \( X^{a}X^{a} \) (colour-blind female), \( X^{A}Y \) (normal vision male), \( X^{a}Y \) (colour-blind male).
- Identify the target phenotype: We are looking for a colour-blind daughter, which corresponds to the genotype \( X^{a}X^{a} \).
- Calculate the probability: There are four equally likely outcomes. Only one outcome (\( X^{a}X^{a} \)) matches the target. Therefore, the probability is \( \frac{1}{4} \).
Units check. Probability is a dimensionless ratio, so no units are required, but the fraction must be expressed in simplest form.
Three quick tests
- Why do X-linked recessive traits appear more frequently in males than in females?
Answer
Males are hemizygous for X-linked genes, meaning they have only one copy. A single recessive allele is sufficient to express the trait, whereas females need two copies. - Can an affected father pass an X-linked recessive trait to his son?
Answer
No. Fathers pass their Y chromosome to their sons, not their X chromosome. Therefore, the son cannot inherit the X-linked allele from his father. - What is the genotype of a carrier female for an X-linked recessive trait?
Answer
The genotype is heterozygous, typically written as \( X^{A}X^{a} \), where \( X^{A} \) is the normal allele and \( X^{a} \) is the recessive disease allele.
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Every lecture and every concept in this unit, with prerequisites and threads.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.
Concepts cited: T-016 · Threads: information, evolution